Level 5 — MasteryThermodynamics (Chemical)

Thermodynamics (Chemical)

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Time limit: 90 minutes Total marks: 60 Instructions: Answer all three questions. Show all working. Use $R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ , $\ln$ / $\log$ as needed. State assumptions explicitly.

Question 1 — First Law, Work, and Heat Capacity Bridge (20 marks)

One mole of an ideal monatomic gas ( $C_{V,m} = \tfrac{3}{2}R$ ) is initially at $T_1 = 300\ \mathrm{K}$ , $P_1 = 5.00\ \mathrm{bar}$ . It undergoes the following two-stage process:

Stage A: Reversible isothermal expansion at $300\ \mathrm{K}$ until the pressure falls to $1.00\ \mathrm{bar}$ .
Stage B: Reversible adiabatic expansion until the volume doubles relative to the volume at the end of Stage A.

(a) Derive from the first law and the ideal-gas equation the general relation $C_{P,m} - C_{V,m} = R$ . State clearly at which point you invoke $H = U + PV$ . (4)

(b) For Stage A, compute $w$ , $q$ , $\Delta U$ , and $\Delta H$ (in J). (5)

(c) For Stage B, derive the final temperature $T_3$ using the adiabatic relation $TV^{\gamma-1}=\text{const}$ , where $\gamma = C_{P,m}/C_{V,m}$ . Then compute $w$ , $q$ , $\Delta U$ , $\Delta H$ for Stage B. (6)

(d) State, with a one-line justification for each, which of $w$ , $q$ , $\Delta U$ , $\Delta H$ are state functions and which are path functions. Then verify that $\Delta U_{\text{total}}$ over A+B could equally be computed by a single-step reversible path between the same endpoints. (5)

Question 2 — Born–Haber Cycle, Hess's Law, and Error Propagation (22 marks)

For solid magnesium chloride $\mathrm{MgCl_2(s)}$ the following standard data (in $\mathrm{kJ\,mol^{-1}}$ ) are given at 298 K:

Quantity	Symbol	Value
Enthalpy of formation of $\mathrm{MgCl_2(s)}$	$\Delta H_f^\circ$	$-641.3$
Sublimation of Mg	$\Delta H_{sub}$	$+147.7$
1st ionization energy of Mg	$IE_1$	$+737.7$
2nd ionization energy of Mg	$IE_2$	$+1450.7$
Bond dissociation of $\mathrm{Cl_2}$	$\Delta H_{diss}$	$+243.0$
Electron affinity of Cl	$EA$	$-349.0$

(a) Construct the full Born–Haber cycle for $\mathrm{MgCl_2(s)}$ and compute the lattice enthalpy $\Delta H_{lat}$ (defined as the enthalpy for $\mathrm{Mg^{2+}(g)} + 2\mathrm{Cl^-(g)} \to \mathrm{MgCl_2(s)}$ ). Show every term with its multiplicity. (9)

(b) The measured enthalpy of solution of $\mathrm{MgCl_2(s)}$ is $\Delta H_{soln}^\circ = -155.0\ \mathrm{kJ\,mol^{-1}}$ , and the sum of hydration enthalpies $\Delta H_{hyd}(\mathrm{Mg^{2+}}) + 2\Delta H_{hyd}(\mathrm{Cl^-})$ is to be found. Using the cycle $\text{lattice} \to \text{gaseous ions} \to \text{aqueous ions}$ , and your $\Delta H_{lat}$ from (a), determine the total hydration enthalpy. State the Hess's law relation you use. (6)

(c) Suppose each of the six input quantities in the table carries an independent uncertainty of $\pm 1.0\%$ of its magnitude. Treating $\Delta H_{lat}$ as a linear combination of the inputs, compute the combined (root-sum-square) absolute uncertainty in $\Delta H_{lat}$ . Comment on which single term dominates and why. (7)

Question 3 — Gibbs, Equilibrium, and Reaction Coupling (18 marks)

Consider the extraction of a metal via the reduction of its oxide by carbon:

$\mathrm{MO(s)} \to \mathrm{M(s)} + \tfrac{1}{2}\mathrm{O_2(g)} \qquad \Delta H_1^\circ = +290.0\ \mathrm{kJ\,mol^{-1}},\ \ \Delta S_1^\circ = +75.0\ \mathrm{J\,K^{-1}mol^{-1}}$

$\mathrm{C(s)} + \tfrac{1}{2}\mathrm{O_2(g)} \to \mathrm{CO(g)} \qquad \Delta H_2^\circ = -110.5\ \mathrm{kJ\,mol^{-1}},\ \ \Delta S_2^\circ = +89.0\ \mathrm{J\,K^{-1}mol^{-1}}$

(a) Show that the uncoupled decomposition (reaction 1 alone) is non-spontaneous at 298 K by computing $\Delta G_1^\circ$ . (3)

(b) Couple reactions 1 and 2 to give the overall reduction $\mathrm{MO(s)}+\mathrm{C(s)}\to\mathrm{M(s)}+\mathrm{CO(g)}$ . Compute $\Delta H^\circ$ , $\Delta S^\circ$ , and $\Delta G^\circ$ of the coupled reaction at 298 K, and determine the threshold temperature above which the coupled reaction becomes spontaneous. (6)

(c) At the threshold temperature $T^*$ , and at $T = T^* + 200\ \mathrm{K}$ , compute the equilibrium constant $K$ using $\Delta G^\circ = -RT\ln K$ (assume $\Delta H^\circ, \Delta S^\circ$ temperature-independent). Comment quantitatively on how $K$ shifts. (6)

(d) Explain, in thermodynamic terms tied to entropy, why coupling with the combustion of carbon is so effective a driving strategy in metallurgy. Reference the sign and magnitude of the relevant $\Delta S$ terms. (3)

Answer keyMark scheme & solutions

Question 1

(a) (4 marks) Start from $H = U + PV$ (invoke definition here, 1 mark). For an ideal gas $PV = nRT$ , so $H = U + nRT$ . At constant $P$ : $C_{P,m} = \left(\frac{\partial H}{\partial T}\right)_P$ , $C_{V,m} = \left(\frac{\partial U}{\partial T}\right)_V$ (1 mark). Differentiating $H = U + nRT$ per mole w.r.t. $T$ : $C_{P,m} = C_{V,m} + R$ (1 mark), since for an ideal gas $U$ depends only on $T$ so $\frac{dU}{dT}$ is the same whatever constraint. Hence $C_{P,m}-C_{V,m}=R$ (1 mark).

(b) (5 marks) Isothermal, $\Delta T=0 \Rightarrow \Delta U = 0$ , $\Delta H = 0$ (ideal gas) (1 mark). $w = -nRT\ln\frac{V_2}{V_1} = -nRT\ln\frac{P_1}{P_2}$ (isothermal ⇒ $V_2/V_1 = P_1/P_2$ ) (1 mark). $w = -(1)(8.314)(300)\ln(5/1) = -8.314\cdot300\cdot1.60944 = -4014\ \mathrm{J}$ (2 marks). $q = -w = +4014\ \mathrm{J}$ (from $\Delta U = q + w = 0$ ) (1 mark).

(c) (6 marks) $\gamma = C_{P,m}/C_{V,m} = (5/2 R)/(3/2 R) = 5/3$ (1 mark). Adiabatic: $T_2 V_2^{\gamma-1} = T_3 V_3^{\gamma-1}$ , with $T_2 = 300$ K, $V_3 = 2V_2$ (1 mark). $T_3 = 300\,(V_2/V_3)^{\gamma-1} = 300\,(1/2)^{2/3} = 300 \times 0.62996 = 188.99\ \mathrm{K}$ (2 marks). Adiabatic: $q = 0$ (1 mark). $\Delta U = nC_{V,m}\Delta T = 1\cdot\tfrac{3}{2}(8.314)(188.99-300) = \tfrac{3}{2}(8.314)(-111.01) = -1384.5\ \mathrm{J}$ ; and $w=\Delta U = -1384.5\ \mathrm{J}$ . $\Delta H = nC_{P,m}\Delta T = \tfrac{5}{2}(8.314)(-111.01) = -2307.5\ \mathrm{J}$ (1 mark).

(d) (5 marks)

$\Delta U$ , $\Delta H$ are state functions — depend only on endpoints (1).
$q$ , $w$ are path functions — differ per route (isothermal vs adiabatic gave different splits) (1).
$\Delta U_{\text{total}} = 0 + (-1384.5) = -1384.5\ \mathrm{J}$ (1). Since $U$ is a state function, a single reversible path from $(T_1=300)$ to $(T_3=189.0)$ gives $\Delta U = nC_{V,m}(T_3 - T_1) = \tfrac32(8.314)(188.99-300)=-1384.5\ \mathrm{J}$ , identical — verifying path independence (2).

Question 2

(a) (9 marks) Born–Haber cycle for $\mathrm{Mg(s)} + \mathrm{Cl_2(g)} \to \mathrm{MgCl_2(s)}$ :

$\Delta H_f^\circ = \Delta H_{sub} + IE_1 + IE_2 + \Delta H_{diss} + 2\,EA + \Delta H_{lat}$

(Note $\Delta H_{diss}$ is for one full $\mathrm{Cl_2}$ → 2 Cl, and $EA$ counted twice for two Cl.) (3 marks for correct cycle & multiplicities)

Solve for $\Delta H_{lat}$ : $\Delta H_{lat} = \Delta H_f^\circ - \Delta H_{sub} - IE_1 - IE_2 - \Delta H_{diss} - 2\,EA$ $= -641.3 - 147.7 - 737.7 - 1450.7 - 243.0 - 2(-349.0)$ $= -641.3 - 147.7 - 737.7 - 1450.7 - 243.0 + 698.0 = -2522.4\ \mathrm{kJ\,mol^{-1}}$ (6 marks; correct arithmetic and sign of exothermic lattice enthalpy)

(b) (6 marks) Hess cycle: $\Delta H_{soln}^\circ = -\Delta H_{lat} + \Delta H_{hyd,total}$ (dissolving = reverse of lattice formation, then hydrate the gaseous ions) (2 marks for relation). $\Delta H_{hyd,total} = \Delta H_{soln}^\circ + \Delta H_{lat} = -155.0 + (-2522.4) = -2677.4\ \mathrm{kJ\,mol^{-1}}$ (4 marks). Strongly exothermic hydration, consistent with small doubly-charged $\mathrm{Mg^{2+}}$ .

(c) (7 marks) Since $\Delta H_{lat}$ is a $\pm1$ linear combination of the six inputs, each term's absolute uncertainty is $1.0\%$ of its magnitude:

| Term | magnitude | $\sigma_i = 0.01\times|x_i|$ | |---|---|---| | $\Delta H_f$ | 641.3 | 6.413 | | $\Delta H_{sub}$ | 147.7 | 1.477 | | $IE_1$ | 737.7 | 7.377 | | $IE_2$ | 1450.7 | 14.507 | | $\Delta H_{diss}$ | 243.0 | 2.430 | | $2\,EA$ | 698.0 | 6.980 |

(The $2\,EA$ term has coefficient 2 but the input EA magnitude is 349 with $\sigma_{EA}=3.49$ ; the 2× term uncertainty = $2\times3.49 = 6.98$ .) (2 marks for setup incl. factor of 2)

RSS: $\sigma = \sqrt{6.413^2 + 1.477^2 + 7.377^2 + 14.507^2 + 2.430^2 + 6.980^2}$ $= \sqrt{41.13 + 2.18 + 54.42 + 210.45 + 5.90 + 48.72} = \sqrt{362.80} = 19.05\ \mathrm{kJ\,mol^{-1}}$ (3 marks)

Dominant term: $IE_2$ (±14.5), because it is both the largest magnitude and carries the largest absolute error; its variance (210.5) is ~58% of the total. (2 marks)

So $\Delta H_{lat} = -2522.4 \pm 19.1\ \mathrm{kJ\,mol^{-1}}$ .

Question 3

(a) (3 marks) $\Delta G_1^\circ = \Delta H_1^\circ - T\Delta S_1^\circ = 290000 - 298(75.0) = 290000 - 22350 = 267650\ \mathrm{J} = +267.65\ \mathrm{kJ}$ . Positive ⇒ non-spontaneous (3).

(b) (6 marks) Coupled: add reactions ( $\tfrac12 O_2$ cancels). $\Delta H^\circ = 290.0 + (-110.5) = +179.5\ \mathrm{kJ}$ (1). $\Delta S^\circ = 75.0 + 89.0 = +164.0\ \mathrm{J\,K^{-1}}$ (1). $\Delta G^\circ(298) = 179500 - 298(164.0) = 179500 - 48872 = +130628\ \mathrm{J} = +130.6\ \mathrm{kJ}$ (still non-spontaneous at 298 K) (2). Threshold: $\Delta G^\circ = 0 \Rightarrow T^* = \Delta H^\circ/\Delta S^\circ = 179500/164.0 = 1094.5\ \mathrm{K}$ (2).

(c) (6 marks) At $T^*$ : $\Delta G^\circ = 0 \Rightarrow \ln K = 0 \Rightarrow K = 1$ (2). At $T = 1294.5\ \mathrm{K}$ : $\Delta G^\circ = 179500 - 1294.5(164.0) = 179500 - 212298 = -32798\ \mathrm{J}$ . $\ln K = -\Delta G^\circ/(RT) = 32798/(8.314\times1294.5) = 32798/10762.6 = 3.048$ . $K = e^{3.048} = 21.1$ (3). $K$ rises from 1 to ~21 over 200 K — reaction becomes clearly product-favored; the entropy-driven $+T\Delta S$ term overtakes the endothermic $\Delta H$ (1).

(d) (3 marks) Coupling with $\mathrm{C+\tfrac12O_2\to CO}$ contributes a large positive $\Delta S_2 = +89\ \mathrm{J\,K^{-1}}$ (net creation of gas: consumes $\tfrac12$ mol $O_2$ but produces 1 mol $CO$ ) and releases heat ( $\Delta H_2<0$ ), lowering the overall $\Delta H$ while raising overall $\Delta S$ . The combined large positive $\Delta S^\circ = 164\ \mathrm{J\,K^{-1}}$ means the $-